Problem: Let $f(x)=\sqrt{x+9}$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $[0,16]$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $16$ (Choice B) B $43$ (Choice C) C $55$ (Choice D) D $7$
Explanation: According to the Mean Value Theorem, there exists a number $c$ in the open interval $(0,16)$ such that $f'(c)$ is equal to the average rate of change of $f$ over the interval: $f'(c)=\dfrac{f(16)-f(0)}{(16)-(0)}$ First, let's find that average rate of change: $\dfrac{f(16)-f(0)}{(16)-(0)}=\dfrac{5-3}{16}={\dfrac{1}{8}}$ Now, let's differentiate $f$ and find the $x$ -value for which $f'(x)={\dfrac{1}{8}}$. $f'(x)=\dfrac{1}{2\sqrt{x+9}}$ The solution of $f'(x)=\dfrac{1}{8}$ is $x=7$. $x=7$ is indeed within the interval $(0,16)$. In conclusion, $c=7$.